Answer
$\frac{x^2+3x+9}{x-1}$.
Work Step by Step
The given expression is
$=\frac{x^3-27}{4x^2-4x}\cdot \frac{4x}{x-3}$
Factor $x^3-27$.
$=x^3-3^3$.
Use the algebraic identity $a^3-b^3=(a-b)(a^2+ab+b^2)$.
$=(x-3)(x^2+3x+9)$
Factor $4x^2-4x$.
$=4x(x-1)$
Substitute all factors into the given expression.
$=\frac{(x-3)(x^2+3x+9)}{4x(x-1)}\cdot \frac{4x}{x-3}$
Cancel common terms.
$=\frac{x^2+3x+9}{x-1}$.
